Answer:
The probability that the student's IQ is at least 140 points is of 55.17%.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
University A: 
a) Select a student at random from university A. Find the probability that the student's IQ is at least 140 points.
This is 1 subtracted by the pvalue of Z when X = 140. So
has a pvalue of 0.4483.
1 - 0.4483 = 0.5517
The probability that the student's IQ is at least 140 points is of 55.17%.
A.) If Lucy sells 1 necklace, her sales would equal to $15.99. Then her profit would be:
Profit = $15.99 - $3.38(1) - $5.57(1)
Profit = $7.04
The fraction of the sale price of the necklace in profit is denoted as x.
15.99x = 7.04
x = 704/1559
b.) This is the same as part (a) but in decimal form. Just simply divide 704 by 1559. The answer is 0.44
c.) If Lucy's sales is $223.86 and each necklace costs $15.99, then the number of necklaces sold is $223.86 ÷ $15.99/necklace = 14 necklaces
Her profit for the 14 necklaces sold would be:
$223.86 - $3.38(14) - $5.57(14) = $98.56
10 - 2x = 120
120 + 10 = 130
130 / 2 = 65
x = 65
2(65) - 10 = 120
The points you found are the vertices of the feasible region. I agree with the first three points you got. However, the last point should be (25/11, 35/11). This point is at the of the intersection of the two lines 8x-y = 15 and 3x+y = 10
So the four vertex points are:
(1,9)
(1,7)
(3,9)
(25/11, 35/11)
Plug each of those points, one at a time, into the objective function z = 7x+2y. The goal is to find the largest value of z
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Plug in (x,y) = (1,9)
z = 7x+2y
z = 7(1)+2(9)
z = 7+18
z = 25
We'll use this value later.
So let's call it A. Let A = 25
Plug in (x,y) = (1,7)
z = 7x+2y
z = 7(1)+2(7)
z = 7+14
z = 21
Call this value B = 21 so we can refer to it later
Plug in (x,y) = (3,9)
z = 7x+2y
z = 7(3)+2(9)
z = 21+18
z = 39
Let C = 39 so we can use it later
Finally, plug in (x,y) = (25/11, 35/11)
z = 7x+2y
z = 7(25/11)+2(35/11)
z = 175/11 + 70/11
z = 245/11
z = 22.2727 which is approximate
Let D = 22.2727
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In summary, we found
A = 25
B = 21
C = 39
D = 22.2727
The value C = 39 is the largest of the four results. This value corresponded to (x,y) = (3,9)
Therefore the max value of z is z = 39 and it happens when (x,y) = (3,9)
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Final Answer: 39
